Page No 14.16:

Question 8:

If α, β are the roots of the equation x2+px+1=0;γ,δ the roots of the equation x2+qx+1=0,then(α-γ)(α+δ)(β-γ)(β+δ)=
(a) q2-p2
(b) p2-q2
(c) p2+q2
(d) none of these.

Answer:

(a) q2-p2
Given: αandβ are the roots of the equation x2+px+1=0.
Also, γandδ are the roots of the equation x2+qx+1=0.
Then, the sum and the product of the roots of the given equation are as follows:α+β=-p1=-pαβ=11=1γ+δ=-q1=-qγδ=11=1

Question 17:

Answer:

It is given that, p and q (p ≠ 0, q ≠ 0) are the roots of the equation x2+px+q=0.

∴Sumofroots=p+q=-p⇒2p+q=0...(1)

Productofroots=pq=q⇒qp-1=0⇒p=1,q=0butq≠0

Now, substituting p = 1 in (1), we get,

2+q=0⇒q=-2

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 14.17:

Question 18:

The set of all values of m for which both the roots of the equation x2-(m+1)x+m+4=0 are real and negative, is
(a) (-∞,-3]∪[5,∞)
(b) [−3, 5]
(c) (−4, −3]
(d) (−3, −1]

Answer:

(c) m∈(-4,-3]

The roots of the quadratic equation x2-(m+1)x+m+4=0 will be real, if its discriminant is greater than or equal to zero.

∴m+12-4m+4≥0⇒m-5m+3≥0⇒m≤-3orm≥5...(1)

It is also given that, the roots of x2-(m+1)x+m+4=0 are negative.
So, the sum of the roots will be negative.

∴ Sum of the roots < 0

⇒m+1<0⇒m<-1...(2)

and product of zeros >0

⇒m+4>0⇒m>-4...(3)

From (1), (2) and (3), we get,

m∈(-4,-3]

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 14.17:

Question 19:

The number of roots of the equation (x+2)(x-5)(x-3)(x+6)=x-2x+4 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(b) 1

(x+2)(x-5)(x-3)(x+6)=(x-2)(x+4)⇒(x2-3x-10)(x+4)=(x2+3x-18)(x-2)⇒x3+4x2-3x2-12x-10x-40=x3-2x2+3x2-6x-18x+36⇒x2-22x-40=x2-24x+36⇒2x=76⇒x=38
Hence, the equation has only 1 root.

Page No 14.17:

Question 20:

If α and β are the roots of 4x2+3x+7=0, then the value of 1α+1β is
(a) 47
(b) -37
(c) 37
(d) -34

Answer:

(b) −3/7

Given equation: 4x2+3x+7=0
Also, α and β are the roots of the equation.

Then, sum of the roots = α+β=-CoefficientofxCoefficientofx2=-34

Product of the roots = αβ=ConstanttermCoefficientofx2=74

∴1α+1β=α+βαβ=-3474=-37

Page No 14.17:

Question 21:

If α, β are the roots of the equation x2+px+q=0then-1α+1β are the roots of the equation
(a) x2-px+q=0
(b) x2+px+q=0
(c) qx2+px+1=0
(d) qx2-px+1=0

Answer:

(d) qx2-px+1=0
Given equation: x2+px+q=0
Also, α and β are the roots of the given equation.
Then, sum of the roots = α+β=-p
Product of the roots = αβ=q
Now, for roots -1α,-1β, we have:
Sum of the roots = -1α-1β=-α+βαβ=--pq=pq
Product of the roots = 1αβ=1q
Hence, the equation involving the roots -1α,-1β is as follows:x2-α+βx+αβ=0

⇒x2-pqx+1q=0⇒qx2-px+1=0

Page No 14.17:

Question 22:

If the difference of the roots of x2-px+q=0 is unity, then
(a) p2+4q=1
(b) p2-4q=1
(c) p2+4q2=(1+2q)2
(d) 4p2+q2=(1+2p)2

Answer:

(b) p2-4q=1
Given equation: x2-px+q=0
Also αandβ are the roots of the equation such that α-β=1.
Sum of the roots = α+β=-CoefficientofxCoefficientofx2=--p1=p